$f(x)=\dfrac{1}{\sqrt{x+1}}$ Find the second degree Maclaurin polynomial of $f$. Choose 1 answer: Choose 1 answer: (Choice A) A $1-\frac{1}{2}x+\frac{3}{8}x^2$ (Choice B) B $ x-\frac{1}{2}x^2+\frac{3}{8}x^3$ (Choice C) C $1+\frac{1}{2}x+\frac{3}{8}x^2$ (Choice D) D $1+\frac{1}{2}x-\frac{3}{8}x^2$
First, find the first two derivatives of $~f(x)=\dfrac{1}{\sqrt{x+1}}=(x+1)^{-1/2}\,$. ${f}\,^\prime(x)=-\dfrac{1}{2}(x+1)^{-3/2}=-\frac{1}{2(x+1)^{3/2}}$ ${f}\,^{\prime\prime}(x)=\dfrac{3}{4}(x+1)^{-5/2}=\dfrac{3}{4(x+1)^{5/2}}$ Then let $~x=0~$ in the original function and these derivatives to get the coefficients for $~{{T}_{2}}\left( x \right)\,$, the second degree Taylor polynomial for $~f\left( x \right)\,$. $ f(0)=1\,; {f}\,^\prime(0)=-\frac{1}{2}\,; {f}\,^{\prime\prime}(0)=\frac{3}{4}$ Use these coefficients in the equation of the Taylor polynomial of degree $~2\,$. ${{T}_{2}}\left( x \right)=f(0)+{f}\,^\prime(0)(x-0)+\frac{{f}\,^{\prime\prime}(0){{(x-0)}^{2}}}{2!}$ Hence, ${{T}_{2}}\left( x \right)=1-\frac{1}{2}x+\frac{3}{8}x^2$